Find a unit vector perpendicular to the plane ABC, where the coordinates of A, B and C are A(3, –1, 2), B(1, –1, –3) and C(4, –3, 1).
Given points A(3, –1, 2), B(1, –1, –3) and C(4, –3, 1)
Let position vectors of the points A, B and C be
,
and
respectively.

We know position vector of a point (x, y, z) is given by
, where
,
and
are unit vectors along X, Y and Z directions.
![]()
![]()
Similarly, we have
and![]()
Plane ABC contains the two vectors
and
.
So, a vector perpendicular to this plane is also perpendicular to both of these vectors.
Recall the vector
is given by
![]()
![]()
![]()
![]()
![]()
Similarly, the vector
is given by
![]()
![]()
![]()
![]()
![]()
We need to find a unit vector perpendicular to
and
.
Recall a vector that is perpendicular to two vectors
and
is

Here, we have (a1, a2, a3) = (–2, 0, –5) and (b1, b2, b3) = (1, –2, –1)

![]()
![]()
![]()
Let the unit vector in the direction of
be
.
We know unit vector in the direction of a vector
is given by
.

Recall the magnitude of the vector
is
![]()
Now, we find
.
![]()
![]()
![]()
So, we have![]()
![]()
Thus, the required unit vector that is perpendicular to plane ABC is
.